How to run multiple select query at once in sequelize using single await

Question

Have to run both codes at a single time. Don't want to wait and then fire next query but wait for both query execution complete.

products = await Product.findAll()
.then(data => {
  return data;
})
.catch(error => {
  //
});

variationProducts = await VariationProduct.findAll()
.then(data => {
  return data;
})
.catch(error => {
  //
});

Answer 1

You may choose

const [ productsPromise, variationProductsPromise ] = await Promise.all([Product.findAll(), VariationProduct.findAll()]);

OR

const [ productsPromise, variationProductsPromise ] = await { Product.findAll(), VariationProduct.findAll()}

Answer 2

try {
    const [products, variationProducts] = await Promise.all([
        Product.findAll(),
        VariationProduct.findAll()
    ]);

    // Do what you need with the result;
}
catch(e) {
    console.error('Problem in getting data', e);
    throw e; // Or do what you want.
}